Polar Moment of Inertia [Kg m ^2]Calculation

[Alfoncito]

Hi everyone? can please anybody help me how can I find Polar Moment of Inertia [Kg m ^2]of the Race car?
thank you all
in advance
Alfonso

[Jersey Tom]

Same way as anything else. Sum of all M*R^2 about an axis of your choosing.

[xpensive]

To be a tad more specific, mass times radius squared applies to a solid cylinder and around its centerline.

Other bodies have different and more complex formulas for mass inertia, that can be found in an engineer's handbook.

[Scotracer]

I wouldn't want to work out the Polar Moment of an entire car. I'm not deriving that expression

[Jersey Tom]

Sum of M*R^2 applies for everything.

In any event. Best way to do it is either in CAD, or a physical test rig. Or, you could get a rough idea from a spreadsheet.

[Ciro Pabón]

In first place, I would like to say thanks, Alfoncito.

You managed to deliver one of the worst best questions (or one of the best worst questions) in the history of this forum. You ask for polar moment of inertia and then you say its units are kg*m2.

Then, you get several responses from a bunch of guys I know well and who are rarely mistaken.

This confirms my very objective opinion about why civil engineer rules: when you study civil engineering, you have to read through so many subjects that in the end you need to understand, instead of memorizing.

Alfoncito, this I will say once:

The polar moment of inertia is a quantity you use to estimate resistance to angular torsion, like in an fixed hinge, when you want to know when will it break by torsion. It’s a measure of how much resistance to twisting around an axis has a section.

In english: picture yourself grabbing the car (real hard!) by the front bumper and your friends trying to break the car by pushing the rear bumper up. The larger the polar moment, the harder the task of your friends.

The area moment of inertia or second moment of inertia, is used to estimate resistance to bending, like in a beam when you want to know how much will it sag under load. It’s a measure of the resistance of a section of a solid to perpendicular loads.

In english: picture yourself (and your friends) bumping up and down on the roof of the car. The larger the second moment of inertia, the harder is to break the car.

The simple moment of inertia is used to estimate resistance to rotation, like in an axle you want to know how fast rotates, given an energy. It’s analog to inertial mass, but for rotation instead of linear displacements.

For the last time, in english: picture yourself grabbing the car by the front bumper, like Superman, and trying to turn the car in the air around your head. The larger the moment of inertia, the harder is to turn the car around.

Polar moment of inertia: m4
Area moment of inertia: m4
Moment of inertia: Kg*m2

If you weren’t able to distinguish them, my old teacher, Otoniel, would have given you an F in Structures.

So, is the moment of Inertia what you’re asking for, or the polar one?

Anyway, my friend, I believe in the units:

For kg*m2, forget about the worksheets by young engineers or the formulas of the old ones (I share Scotracer fear of the huge formula, the integrals and the such that you'd had to find for a race car, including the density, for the love of Pete!).

So, the simple answer is:

Use Autocad MASSPROP function as any sane civil engineer does since his infancy (the icon in the Properties toolbar that displays a solid and a little rule under it).

Draw the car in Autocad and calculate MASSPROP for the XY plane. If you want the POLAR moment, remember that the polar in Z is equal to the second moment in X plus the second moment in Y (if I remember correctly).

Jz = Ix + Iy

If you have to answer it without Autocad (like in an exam or worse), then read the wiki article on moments of inertia, as any sane student does.

Basically, as JTom insinuates, you divide the car or object in small squares and multiply its tiny area (or mass, depending on what you want to calculate) by the distance to the axis of rotation, plane of bending or point of torsion.

BTW, in Autocad, draw the area in the XY plane of the UCS (the funny axis “thingie” in the lower left corner of the screen).

In English: draw the object “flat”.

If you have a 3D drawing or if you want to calculate the number for different axis, rotate your car in the UCS plane accordingly and take a section of it.

Now, I wonder if I made a mistake… I’m sure somebody will notice, but what the heck, he will have to read the damn wiki article… and come back to explain if Otoniel missed me.

Besides, I haven't used the MASSPROP function that much, perhaps somebody knows how to calculate the different moments of inertia from it (the function gives you the POLAR moment of inertia, if I'm not mistaken). I vaguely remember that you divide by the radius of gyration squared (maybe) to get what I've called the simple moment of inertia, but I'm not sure.

You know, I'm civil, I use it to get the properties of a beam under bending loads and the such, I don't rotate the beams that much (but I can distinguis both cases, unlike mechanicals ).

[n smikle]

Civil engineers are lame.. Basically watered down "mech men"

Answered like a real "Civ man"


A mech-man would answer it different.
More like this:

Put the car into basic shapes and get the Moment of Inertia equations for each of those shapes then combine them.. (Parallel axis theorem etc).. It is a tall order because of complex object like the engine and the body of the car. which have many sub-parts with different densities.

Because of the complexity I think it's best to find it by computer model or by experiment.

Experimental example (just out of my head.. I don't know if people do this):

Put the car on some rotational rig and spin it around..accelerate it or spin it then slow it down with a brake. Then measure the angular acceleration/deceleration and the accelerating torque/braking torque.

Then using

Torque = I x alpha

you can get the Moment of Inertia around the rotation axis.

[xpensive]

Beautiful! Take it from an old-geezer, you're my kind of guy smikle!

[DaveW]

It is possible to derive sprung mass pitch moment of inertia directly from a four post rig test.

Not sure about other rigs, but it will be computed automatically when using either of Multimatic's rigs (one in Thetford, UK, the other in Toronto, Canada) when it is used to help to set-up the suspension of any vehicle (road or race).

It is relatively simple to estimate overall vehicle yaw inertia from derived inertia parameters (sprung & unsprung masses, longitudinal c.g. position & sprung mass pitch inertia).

Such "basic" information can sometimes be surprisingly useful. For example, one customer discovered that installing ballast in the right place made success ballast a benefit, rather than a handicap (lap times were better with success ballast).

[n smikle]

Beautiful, coming from an old-geezer engineer, you're my kind of guy smikle!

I sound like and old geezer? lol I just finished school last year!

[xpensive]

My last post was poorly written, I'm the old-geezer around here.

[fletchy]

Hi, I'm doing a similar thing and I can't use CAD because some of the part densities aren't accurate. I need to calculate the mass moment of inertia of an entire vehicle. From what I have gathered and can remember from my dynamics of rigid body module last year;

I can sum all of the inertias of each part by using their mass and centre of gravity using;

I = m(mass of part) * r(radial distance from reference axis)^2

However some of the parts have quite significant inertias of their own with respect to the entire vehicle (eg, the chassis) so I need to add these in. Therefore, due to the parallel axis theorem (http://emweb.unl.edu/NEGAHBAN/EM223/note19/note19.htm), I can add the above calculation to the calculated inertia for the part, which I can get from CAD for the significant parts.

I(part relative to origin) = I(part relative to it's own COG) + m(mass) * r(radial dist..)^2

Then sum all of that up to give the whole vehicle.

What I do still need to know is; does the reference axis matter, or will it be cancelled out eventually as long as I use the same one for each part?

I need the mass moment of inertia about the COG I think, but I'm calculating the COG in parallel so won't be able to do that from the beginning. I was thinking about using the SAE origin (mid track front axle, on the floor) as a reference point. I have an idea that it may be the parallel axis theorem again?

Do I need to then subtract the inertia calculated from the mass of the vehicle multiplied by the radial distance from the COG of the reference axis?

PS: To our (less than) civil friend, calling the 'simple' moment of inertia, the mass moment of inertia is a lot clearer to distinguish the difference.

[speedsense]

So... how do I calculate my driver's moment of inertia? Not his body, but his brain? And how do I calculate the moment of inertia of the rubber band effect of the tires, from which all the equations here revolve? Look at one thing and the other aspect changes the math..the chaos is in the tires and the drivers hands and feet... and lastly the one thing nobody here can make math from, your driver's thoughts...

[xpensive]

And Ingmar Bergman made booring movies.

[speedsense]

And Ingmar Bergman made booring movies.

Ingmar Bergman? that's not old, that's ancient, wasn't she buried in a pyramid?